Answer:
To determine the mass of Al that was needed to produce 47.5 L of H2 gas at STP, we can use the stoichiometry of the balanced chemical equation to calculate the amount of Al required.
First, we need to convert the volume of H2 gas at STP to the corresponding amount in moles. At STP, 1 mole of gas occupies 22.4 L of volume. Therefore, 47.5 L of H2 gas is equal to 2.12 moles of H2 gas.
From the balanced chemical equation, we can see that 3 moles of H2 gas are produced for every 2 moles of Al consumed. Therefore, the amount of Al needed to produce 2.12 moles of H2 gas can be calculated as:
2.12 moles H2 x (2 moles Al / 3 moles H2) = 1.41 moles Al
Finally, we can use the molar mass of Al to convert moles of Al to its mass in grams. The molar mass of Al is 26.98 g/mol. Therefore, the mass of Al required can be calculated as:
1.41 moles Al x 26.98 g/mol = 38.1 g Al (rounded to one decimal place)
Therefore, approximately 38.1 grams of Al were needed to react with excess HCl to produce 47.5 L of H2 gas at STP.
Step-by-step explanation:
In this balanced chemical equation, 2 moles of Al react with 6 moles of HCl to produce 2 moles of AlCl3 and 3 moles of H2. We are given the volume of H2 gas produced at STP and asked to calculate the mass of Al needed to react with excess HCl to produce this amount of H2 gas. To solve this problem, we used stoichiometry to relate the amount of H2 gas produced to the amount of Al required, and then used the molar mass of Al to convert from moles of Al to its mass in grams.
Hope this helps!!